I am seeking a clarification to this question on almost convergence in Sobolev spaces. The essence of the answer there was to prove that if a sequence $(u_n) \rightharpoonup u$ in $H_0^1({\mathbb{R}}^N)$, then upto subsequences, $u_k(x) \rightharpoonup u(x)$ a.e on $\mathbb{R}^N$. Since the embedding theorems cannot be directly applied for $\mathbb{R}^N$, the approach was to consider open balls of increasing radii and somehow prove the statement. This method was also given here.

I'm not sure if this method or the reasoning behind it works. If we define $\Omega_k = B(0,k)$, we can find, for each $k$, a subsequence $\{u_n^{(k)}\}_n$ that converges a.e to $u(x)$ on $\Omega_k$, but it is unclear how one would take the limit.

More precisely, for any $\varepsilon >0 $, let $N_k$ be an integer such that $n > N_k \implies |u_n^{(k)}(x) - u(x)| < \varepsilon, \text{ a.e on } \Omega_k$. To prove the statement, we need to find a subsequence $(u_n)$ such that there always exists an $N$ with the property $n > N \implies |u_n(x) - u(x)| < \varepsilon$. I don't see how one can prove that, since $N$ can "escape to infinity". Any help is much appreciated.


1 Answer 1


I believe you can use a "diagonal argument".

First, you extract a subsequence $\{ u_n^{(1)} \}$ from $\{ u_n \}$ such that $u_n^{(1)}$ converges pointwise inside $\Omega_1$.

Then, from this subsequence, you extract $\{ u_n^{(2)} \}$ which converges pointwise inside $\Omega_2$.

Iteratively, you defined the subsequence $\{ u_n^{(k+1)} \}$ as being extracted from $\{ u_n^{(k)} \}$ and converging pointwise inside $\Omega_{k+1}$.

Eventually, you define, say, $v_n := u_n^{(n)}$ (hence the "diagonal" name for the argument).

Now take any $x \in \mathbb{R}^N$. For $k$ large enough $x \in \Omega_k$. And for $n \geq k$ large enough, $v_n$ is an element of a subsequence of $\{ u^{(k)} \}$, so $v_n(x) \to u(x)$.

  • $\begingroup$ Can you please clarify how one can always extract $u_n^{(2)}$ from $u_n^{(1)}$? I don't intuitively see how a sequence converging on a smaller domain necessarily has a convergent subsequence on a larger domain. (I'm an undergrad and this is my first contact with Functional analysis, so you may need to excuse the pedantry :) ) $\endgroup$ Apr 26, 2023 at 9:20
  • $\begingroup$ Take a bounded sequence in $H^1(\mathbb{R}^N)$. For any compact $K \subset \mathbb{R}^N$, by the Rellich-Kondrachov compact embedding theorem, it has a subsequence strongly converging in $L^1(K)$. In particular, this subsequence converges pointwise almost everwhere in $K$. $\endgroup$
    – cs89
    Apr 26, 2023 at 9:29
  • $\begingroup$ Now once you have constructed $u_n^{(1)}$ to converge pointwise in $\Omega_1$, this subsequence is still bounded in $H^1(\mathbb{R}^N)$. So you can apply the previous comment to extract from itself a subsequence designed for $\Omega_2$. And so on. $\endgroup$
    – cs89
    Apr 26, 2023 at 9:30
  • $\begingroup$ Got it, thanks for the clarification. $\endgroup$ Apr 26, 2023 at 9:34

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